Suppose $A\in \mathbb R^{m\times n}$. I'm supposed to prove $\exists x$ such that $\|Ax-b\|$ in minimal and this $x$ is unique if $A$ is invertible, in which case I also need to exhibit a formula.
I was thinking that since everything is finite dimensional, the image of $A$ (as an operator) is closed, so we can orthogonaly project $b$ onto it and obtain the solution. However, this seems like the only solution without assuming the invertibility of $A$... For a formula, I think the usual projection formula works, and it also doesn't involve the invertibility of $A$..
What am I missing?
Let $V\subseteq\mathbb{R}^n$ be the image of $A$, and let $W\subseteq \mathbb{R}^n$ be an orthogonal complement thereof. Then we can write uniquely $b=b_v+b_w$, with $b_v\in V$ and $b_w\in W$. Then anything from $\mathbb{R}^m$ mapping to $b_v$ will do the trick.
And for the uniqueness when $A$ is invertible (or better, when $\ker{A}=0$) follows immediately.